2002 AMC 10P Problems/Problem 3

Problem

Mary typed a six-digit number, but the two $1$ s she typed didn't show. What appeared was $2002.$ How many different six-digit numbers could she have typed?

$\text{(A) }4 \qquad \text{(B) }8 \qquad \text{(C) }10 \qquad \text{(D) }15 \qquad \text{(E) }20$

Solution 1

This is equivalent to ${5 \choose 2} + 5$ since we are choosing $5$ spots (three in between $2002$ and two to the left and right with two $1$ s. We also have to take into account "double $1$ s," namely, $112002, 211002, 201102, 200112,$ and $200211$ in our $5$ spots.

Thus, our answer is ${5 \choose 2} + 5 = \boxed{\textbf{(D) } 15}.$

Solution 2

We can split this into a little bit of casework which is easy to do in our head.

Case 1: The first $1$ is ahead of the first $2.$ Then the second $1$ has $5$ places.

Case 2: The first $1$ is below the first $2.$ Then the second $1$ has $4$ places.

$\dots$

This continues as the answer comes to

$5 + 4 + 3 + 2 + 1 = \frac{5(6)}{2}=3(5)=15.$

Thus, our answer is $\boxed{\textbf{(D) } 15}.$

Solution 3

We can just count the cases directly since there are so little.

Thus, our answer is $\boxed{\textbf{(D) } 15}.$

2002 AMC 10P (Problems • Answer Key • Resources)
Preceded by Problem 2	Followed by Problem 4
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

Page

Toolbox

Search

2002 AMC 10P Problems/Problem 3

Contents

Problem

Solution 1

Solution 2

Solution 3

See also